95=-5t^2-4t+120

Solution for 95=-5t^2-4t+120 equation:

95=-5t^2-4t+120

We move all terms to the left:

95-(-5t^2-4t+120)=0

We get rid of parentheses

5t^2+4t-120+95=0

We add all the numbers together, and all the variables

5t^2+4t-25=0

a = 5; b = 4; c = -25;
Δ = b2-4ac
Δ = 42-4·5·(-25)
Δ = 516
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{516}=\sqrt{4*129}=\sqrt{4}*\sqrt{129}=2\sqrt{129}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{129}}{2*5}=\frac{-4-2\sqrt{129}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{129}}{2*5}=\frac{-4+2\sqrt{129}}{10}
$